How do you solve $ln(3x+1)-ln(5+x)=ln2$ ?

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Answer 1 · 2015-10-01T08:48:02

$x=9$

$ln(3x+1)-ln(5+x)=ln(2)=>$ using laws of logs:

$ln[(3x+1)/(x+5)]=ln(2)=>$ if $ln(A)=ln(B)hArrA=B$ :

$(3x+1)/(x+5)=2=>$ multiply by $(x+5)$ :

$3x+1=2(x+5)=>$ expand right side:

$3x+1=2x+10=>$ subtract $-2x$ and 1 from both sides:

$3x-2x=10-1=>$ simplify:

$x=9$

How do you solve ln(3x+1)-ln(5+x)=ln2ln(3x+1)-ln(5+x)=ln2?